Problem: Let $f : \mathbb{R} \to \mathbb{R}$ be a function such that
\[f(x^2 + yf(z)) = xf(x) + zf(y)\]for all real numbers $x,$ $y,$ and $z.$

Let $n$ be the number of possible values of $f(5),$ and let $s$ be the sum of all possible values of $f(5).$  Find $n \times s.$
Answer: Setting $x = y = 0,$ we get
\[f(0) = zf(0)\]for all $z,$ so $f(0) = 0.$

Setting $y = 0,$ we get
\[f(x^2) = xf(x)\]for all $x.$

Setting $x = 0,$ we get
\[f(yf(z)) = zf(y).\]In particular, for $y = 1,$ $f(f(z)) = zf(1).$

Since $f(x^2) = xf(x),$
\[f(f(x^2)) = f(xf(x)).\]But $f(f(x^2)) = x^2 f(1)$ and $f(xf(x)) = xf(x),$ so
\[x^2 f(1) = xf(x).\]Then for $x \neq 0,$ $f(x) = f(1) x.$  Since $f(0) = 0,$
\[f(x) = f(1) x\]for all $x.$

Let $c = f(1),$ so $f(x) = cx.$  Substituting into the given equation, we get
\[cx^2 + c^2 yz = cx^2 + cyz.\]For this to hold for all $x,$ $y,$ and $z,$ we must have $c^2 = c,$ so $c = 0$ or $c = 1.$

Thus, the solutions are $f(x) = 0$ and $f(x) = x.$  This means $n = 2$ and $s = 0 + 5,$ so $n \times s = \boxed{10}.$